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A function $f:\mathbb{R}^n\to\mathbb{R}$ is called coercive if $$ \lim_{\|x\|\to\infty}f(x)=\infty. $$ To show that $f$ is coercive, we need to prove that for every sequence $\{x_n\}$ with $\|x_n\|\to\infty$, it holds that $f(x_n)\to\infty$.

My question: Is it sufficient to check that $f$ is coercive along every line? That is, does the condition $$ \lim_{t\to\infty}f(x+td)=\infty,\quad\forall\,x\in\mathbb{R}^n,\quad\forall\,d\in\mathbb{R}^n\backslash\{0\} $$ imply that $f$ is coercive? If not, what additional assumptions (e.g., convexity and/or $C^1$) can we impose on $f$ to make this true?

Rodrigo de Azevedo
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Kittayo
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  • I feel like (without continuity at least) you should be able to construct a counterexample that is equal to $1$ on, say, the curve $y = e^x$ (for dimension one), and "extend it" to a coercive along every line function, but do take my comment with a grain of salt. – Bruno B Mar 29 '23 at 18:34
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    Counterexample here: https://math.stackexchange.com/a/2093339/42969 – Martin R Mar 29 '23 at 18:38

1 Answers1

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In general, this is not true. However, if we assume $f$ to be continuous and quasi-convex, then it is correct.

Suppose $f$ is not coercive, then there must exist an unbounded sublevel set $S_a=\{x:f(x)\leq a\}$ of $f$ (see this link). Since $f$ is quasi-convex, $S_a$ is convex. For an unbounded convex set, its recession cone $R_{S_a}$ must contain a nonzero element $d\ne 0$ (see Proposition 1.4.2 of this book). That's to say, there is $x\in S_a$ such that $x+td\in S_a$ for all $t\geq 0$. Recall that $f$ is coercive along every line, so $f(x+td)\to \infty$ as $t\to\infty$. However, that $x+td\in S_a$ implies $f(x+td)\leq a$ for all $t\geq 0$. This leads to a contradiction. Therefore, $f$ must be coercive.

William
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